
By Proposition REF  and Theorem REF  we have \(|G_{L_{pq}}|=2\cdot h_2(L_{pq}^*)=2^{m+1}\) . Assume that \(p\)  and \(q\)  verify conditions (REF ).
According to [1]}, the group \(G_{k_{pq}}\)  is quaternion or semidihedral.
By Galois theory we have
\(Gal(L_{pq}^{(2)}/k_{pq})\simeq Gal(L_{pq}/k_{pq})\times G_{L_{pq}}.\) 
Thus \(G_{L_{pq}}\)  is a subgroup of \(G_{k_{pq}}\)  of index 2. Therefore \(G_{L_{pq}}\)  is dihedral or quaternion. Since a FSU of \(L_{pq}\)  is \(\lbrace \varepsilon _{2}, \varepsilon _{pq},\sqrt{\varepsilon _{pq}\varepsilon _{2pq}}\rbrace \)  (cf.[2]}), then by [3]}, Theorem REF  and Lemma REF , the number of classes of \(\mathrm {Cl}_2(L_{pq})\)  which capitulate in \(L_{pq}^*\)  is \([L_{pq}^*:L_{pq}][E_{L_{pq}}:N_{L_{pq}^*/L_{pq}}(E_{L_{pq}^*})]=2\cdot 1=2\) .
So from Table REF , we deduce that \(G_{L_{pq}}\)  can not be dihedral. Hence \(G_{L_{pq}}\)  is quaternion.
Suppose now that \(p\)  and \(q\)  verify the condition \( {(\ref {cond 2})} \) . As previously we show that \(G_{L_{pq}}\)  is a subgroup of \(G_{k_{pq}}\)  of index 2 and by [1]} \(G_{k_{pq}}\)  is quaternion. So \(G_{L_{pq}}\)  is quaternion.

Let \(k^{\prime }=\mathbb {Q}(\sqrt{-1}, \sqrt{2p^{\prime }}) \) . As \(L_{p^{\prime }}=\mathbb {Q}(\sqrt{-1}, \sqrt{ p^{\prime }}, \sqrt{ 2})\)  is the genus field of \(k^{\prime }\) , so
by [5]}, the 2-class group of \(k^{\prime }\)  is of type \((2, 4)\) . Hence
[6]} implies that
the Hilbet 2-class field tower of \(k\)  terminates at the first step. Therefore, the Hilbert 2-class field tower of its genus field \(L_{p^{\prime }}\)  terminates at the first step. So the result.

